Kitonum

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11 years, 188 days

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These are replies submitted by Kitonum

@tiaranain  I fixed your syntax errors below:

SYS := [D(x)(t)=x(t)-x(t)*y(t)-x(t)*z(t),
D(y)(t)=-0.4*y(t)+x(t)*y(t),
D(z)(t)=-0.24*z(t)+x(t)*z(t)];

RHS := eval( map( rhs, SYS ), [x(t)=x,y(t)=y,z(t)=z] );
                             
plots:-fieldplot3d( RHS, x=-1..1, y=-1..1, z=-1..1, grid=[5,5,5], axes=boxed );

 

@Earl 

LongSum:=proc(f,n0,n)
uses InertForm;
Display(`%+`(seq(f(k),k=1..n0))+`...`+f(n));
end proc:

Examples:
LongSum(n->x[n], 2, n);
LongSum(n->n/(n+1), 3, k);
LongSum(n->n/(n+1), 3, 100);

@minhthien2016  The code shows that this property is also true for a non-convex quadrilateral (in this example). If you still need a convex one, take the point symmetric to DD with respect to the AC. Then you get a convex quadrilateral with the same side lengths.

@tomleslie 
 

isEquiv(Pi*x+y=Pi, x+y/Pi=1); 
isEquiv(x-I*y=I, I*x+y=-1);
isEquiv(sin(x)^2+cos(x)^2=1, 1=1);


Here is my version, which works for these examples:

g:= z-> simplify( (lhs(z)-rhs(z))):
isEquiv1:= (x,y)-> is(simplify(g(x)-g(y))=0 or type(simplify(evalc(g(x)/g(y))),complexcons)):

 

@Christian Wolinski 

'cos'(Pi/2-op(1,convert((1/2)*sqrt(3)*sin(y)+(1/2)*cos(y), phaseamp, y)));

 

@Caneee315 A parallelogram rule is used to add two vectors. See an example below. The animation in this example can be done analogously to the example above:

restart;
with(plots):	
u, v:=[1, 2, 3], [-1, 2, 3]:
U:=arrow(u, color=blue, width=0.1):
V:=arrow(v, color=blue, width=0.1):
W:=arrow(u+v, color=red, width=0.1):
L1:=plottools:-line(u,u+v, linestyle=2, thickness=3, color=black):
L2:=plottools:-line(v,u+v, linestyle=2, thickness=3, color=black):
display(U,V,L1,L2,W, axes=normal, orientation=[-115,15,120]);

                     

 

@Jayaprakash J  Your function depends on 2 variables  x  and  alpha . Therefore, if you want to build it as a function of alpha, then you need to fix the value of  x . Or plot the function of two variables using the plot3d command:

f:=x->-0.3075000000e-1*x^2-0.2145927617e-3*x^6-0.2692226562e-4*x^4+0.3750551906e-2*x^5+0.1050625000e-2*x^3-0.5266789021e-5*x^7+0.1082238239e-4*x^9-6.80130209*10^(-7)*x^6*sin(alpha)+0.2932994791e-5*x^6*cos(alpha)+0.4375000000e-2*x^4*cos(alpha)-0.7990625000e-4*x^5*sin(alpha)-0.1704062500e-3*x^5*cos(alpha)-3.907366066*10^(-7)*x^8*cos(alpha)^2-0.5000000000e-1*x^3*sin(alpha)+0.3375000000e-2*x^4*sin(alpha)-0.2231049199e-3*x^7*sin(alpha)+0.1723809524e-4*x^7*cos(alpha)^2-0.5000000000e-1*x^3*cos(alpha)+7.912017277*10^(-7)*x^8+0.4464285714e-3*x^7*cos(alpha)^3-0.6695959034e-3*x^7*cos(alpha)+0.4337937076e-4*x^9*cos(alpha)^2-0.982142856e-5*x^8*cos(alpha)^3+0.3392338088e-4*x^8*cos(alpha)+0.4906693567e-5*x^8*sin(alpha)+0.4412678315e-5*x^9*cos(alpha)+0.1833880452e-5*x^9*sin(alpha)+6.885206662*10^(-7)*x^10-0.1833333334e-3*x^6*cos(alpha)^2+.6*x+1:
G:=D(f)(x);
plot(eval(G, x=1), alpha=-Pi..Pi);
plot3d(G, alpha=-Pi..Pi, x=0..1);

 

@Carl Love  Yes of course.

The record  12^6^5   is ambiguous. It can be understood as  (12^6)^5  or as  12^(6^5) .

@Carl Love  To prevent extra zeros in output  [8500., [F = HFloat(50.00000000000008), K = HFloat(149.99999999999991)]], we can solve symbolically using the simplex:-maximize  command:

simplex:-maximize(P.V, {seq(C.V <=~ R), seq(V >=~ 0)});
eval(P.V, %);
                        
 {F = 50, K = 150}
                                 8500
 

 

@Jayaprakash J  In Maple 2018.2, the code works correctly. You probably have an earlier version of Maple. Try the following option:

restart;
f:=1+B*x-(1/12)*B*x^3+0.1666666667e-4*B^3*x^3-4.166666667*10^(-8)*B^4*x^4+(1/160)*B*x^5+8.333333333*10^(-11)*B^5*x^5-0.5000000000e-2*B^2*x^2+0.1666666667e-4*B*x^3*C^2-4.166666667*10^(-8)*B*x^4*C^3+8.333333333*10^(-11)*B*x^5*C^4-0.5000000000e-2*B*C*x^2+0.3333333333e-4*B^2*x^3*C-1.250000000*10^(-7)*B^3*x^4*C-1.250000000*10^(-7)*B^2*x^4*C^2+3.333333333*10^(-10)*B^4*x^5*C+5.000000000*10^(-10)*B^3*x^5*C^2+3.333333333*10^(-10)*B^2*x^5*C^3+0.7291666667e-3*B*x^4*C-0.3333333333e-5*B*x^5*C^2+0.6250000000e-3*B^2*x^4-0.2083333333e-5*B^3*x^5-0.5416666667e-5*B^2*x^5*C:
Explore(plot(f, x=-4..4, view=-10..10), parameters=[B=-5...5., C=-5...5.]);

 

@vv  I use Maple 2018.2  64 bit

@vv  I use Maple 2018.2

@Carl Love  Of course you are right. But maybe my simpler solution for OP will be clearer. 

@minhthien2016  It can be clearly seen from the figure in your post that the segment connecting the origin with the center of the pink circle is perpendicular to the plane of this circle. Therefore, the circle in your example cannot be built by the command  \tdplotCsDrawCircle[style]{r}{alpha}{beta}{epsilon}

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